Find the points of intersection of the circles: x^2 + y^2-2x-2y-2=0 and x^2+y^2+2x+2y-2=0. Draw the circles
+2094 It has a pretty reasonable explanation... Sorry I cannot do this.
Maybe this will help more.
https://mathworld.wolfram.com/Circle-CircleIntersection.html
+128474 x^2 + y^2-2x-2y-2=0 and x^2+y^2+2x+2y-2=0
Setting these equal we get that
x^2 + y^2-2x-2y-2 = x^2+y^2+2x+2y-2 simplify
-2x - 2y = 2x + 2y
-4x = 4 y
-x = y
Sub this into eithe equation for y
x^2 + (-x)^2 - 2x - 2(-x) - 2 = 0
x^2 + x^2 - 2x + 2x - 2 = 0
2x^2 - 2 = 0
2x^2 = 2
x^2 = 1 take both roots and we have that
x = 1 and x = - 1
And when x =1, y =-1
And when x = -1, then y = 1
So the points of intersection are ( -1, 1) and (1 , -1)
See the graphs here : https://www.desmos.com/calculator/rjkmzcqniv
